fork() in for() loop

Question

I'm trying to make a homework assignment where I have to use fork() but I don't know why I can't stop my forks after running them through my for loop:

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>

int main(int argc, char *argv[]){
    int limit = argc/2;
    if(argc%2 == 0){

            perror("the number of arguments given must pe even!");
            exit(1);
    }

    int i;
    for(i=0; i<=limit; i++){
        if(fork()==-1){
            perror("childes couldn't be created\n");
            exit(1);
        }
        if(fork()==0){
            printf("fork: %d \n",i);
            exit(1);
        }
        wait(0);
    }


    printf("exiting...\n");
    return 0;
}

Output:

warzaru@ubuntu:~/OS/UprocH$ ./exe a b c d
fork: 0 
fork: 0 
fork: 1 
fork: 1 
fork: 1 
fork: 2 
fork: 2 
fork: 1 
fork: 2 
exiting...
exiting...
fork: 2 
exiting...
exiting...
fork: 2 
exiting...
exiting...
exiting...
warzaru@ubuntu:~/OS/UprocH$ fork: 2 
fork: 2 
fork: 2 
exiting...

Answer 1

You could use execl() or some other variant of exec, this ensures that the child being created does not inherit anything from the parent. This would allow you to create children with a for loop without going into an exponential child explosion. Modify the arguments of the execl function suitably and you can create 3 processes like:

main()
{
    int i,pid[3];
    for(i=0;i<3;i++)
    {
        pid[i]=fork();
        if (pid[i]==0)
            execl("path",0,0);
    }

    sleep(2);
}

Answer 2

Daniel Fischer forced me to provide an answer.

Change:

if(fork()==-1){} // first fork
if(fork()==0){}  // second fork

To:

pid = fork();
if(pid == -1)
{
    ... //failed
}
else if(pid == 0)
{
    ... //success
}

Or use a switch statement:

switch(fork())
{
    case -1:
        ... // failed
        break;
    case 0:
        ... // success
        break;
}